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avatar+287 

Find the only real number that can be expressed in the form 

 

\((a + bi)^3 - 107i\)

 

where \(i^2=-1\), and \(a\) and \(b\) are positive integers.

 Dec 30, 2019
 #1
avatar+109563 
+2

(a + bi)^3  =

 

a^3  + 3a^2 (bi)  +  3a (bi)^2 + (bi)^3

 

a^3  + 3a^2bi  - 3ab^2 - bi

 

We need to  have

 

3a^2bi  - bi  =   107i

 

bi ( 3a^2 - 1)  =  107i                divide both sides by i

 

b ( 3a^2 - 1)  = 107

 

Since 107 is prime.....then....

 

Let  b   = 1

 

So

 

3a^2 - 1 = 107

3a^2  = 108

a^2 = 36

a = 6

 

So

 

(6 + i)^3    - 107i  =

 

6^3  + 3(6)^2 i   + 3(6)(i)^2  - i   - 107i   =

 

216 + 108i  - 18 - i  - 107i   =

 

(216 - 18) +  (108i - 108i)  =

 

198  =  the real number  

 

 

cool cool cool

 Dec 30, 2019
 #2
avatar+287 
+1

I actually figured it out a while after I posted this question haha. My process was the exact same as yours, though

 Dec 30, 2019

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